Find an equation of the line L, where L is perpendicular to y=3x and passes through the point (1,3)
I know the slope would be -1\3x
but in my answer keyy it says the answer is -1\3x + 10/3 I am not sure how they got the +10\3 part
thanks
You have a slope and a point, so
y-3 = -1/3 (x-1)
Now all should be clear.
To find the equation of a line that is perpendicular to y = 3x and passes through the point (1,3), we need to determine the equation in the form y = mx + b, where m is the slope of the line and b is the y-intercept.
The line y = 3x has a slope of 3, which means the perpendicular line will have a slope that is the negative reciprocal of 3. In other words, the perpendicular line will have a slope of -1/3.
We can use the point-slope form of the equation to find the equation of the perpendicular line:
y - y1 = m(x - x1),
where (x1, y1) is the point (1,3) and m is the slope (-1/3).
Plugging in these values, we get:
y - 3 = (-1/3)(x - 1).
Next, we simplify the equation by distributing -1/3 on the right side:
y - 3 = (-1/3)x + 1/3.
To isolate y, we can add 3 to both sides:
y = (-1/3)x + 1/3 + 3.
Combining the fractions on the right side:
y = (-1/3)x + 1/3 + 9/3
y = (-1/3)x + 10/3.
Therefore, the equation of the line L that is perpendicular to y = 3x and passes through the point (1,3) is y = (-1/3)x + 10/3.